Problem: $\begin{aligned} &g(x)=\dfrac{3x-5}{x+1} \\\\ &h(y)=\sqrt{1-3y} \end{aligned}$ $h(g(0))=$
Explanation: When evaluating composite functions, we work our way inside out. To evaluate $h(g(0))$, let's first evaluate $g(0)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $g({0})$. $\begin{aligned}g(x)&=\dfrac{3x-5}{x+1}\\\\ g({0})&=\dfrac{3({0})-5}{({0})+1} ~~~~~~~~~~\text{Plug in }x={0}\\\\ &=\dfrac{-5}{1}\\\\ &={-5}\end{aligned}$ We now know that $h(g({0}))$ is the same as $h({-5})$ because $g({0}) = {-5}$. Let's evaluate $h({-5})$. $\begin{aligned}h(y)&=\sqrt{1-3y}\\\\ h({{-5}})&=\sqrt{1-3({-5})}~~~~~~~~~~\text{Plug in }y={-5}\\\\ &=\sqrt{1-(-15)}\\\\ &=\sqrt{16}\\\\ &=4\end{aligned}$ The answer: $h(g(0))= 4$